1 Answer
1

I assume that you mean $\vec{r}(0)=(1,8,0)$ and $\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}=(4,2,0)$ and that $r=|\vec{r}|$, that is,
$$
r=\sqrt{65}\tag{1}
$$
Note that $r^2=\vec{r}\cdot\vec{r}$ so that $r\frac{\mathrm{d}r}{\mathrm{d}t}=\vec{r}\cdot\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$. Thus, $\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{1}{\sqrt{65}}(1,8,0)\cdot(4,2,0)$. That is,
$$
\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{4}{13}\sqrt{65}\tag{2}
$$
We also have $r^2\frac{\mathrm{d}\theta}{\mathrm{d}t}\vec{n}=\vec{r}\times\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$, where $\vec{n}$ is the unit normal to the plane of motion. Thus, $\frac{\mathrm{d}\theta}{\mathrm{d}t}\vec{n}=\frac{1}{65}(1,8,0)\times(4,2,0)=-\frac{6}{13}(0,0,1)$. That is,
$$
\frac{\mathrm{d}\theta}{\mathrm{d}t}=-\frac{6}{13}\tag{3}
$$
I further assume that $\theta(1,0,0)=0$. Then, $\theta(1,8,0)=\arctan(8)$. That is,
$$
\theta=\arctan(8)\tag{4}
$$

Thankyou for this. However, i have a query. The second line of your computation after r^2=r.r, so that ........ Thus. I don't understand where the differential equations before the word thus come from?
–
EudenMar 29 '12 at 15:51

Take the derivative of each side of the previous equation with respect to $t$ and divide by $2$. Note that $\vec{r}$ is a vector and $r$ is a scalar ($r=|\vec{r}|$).
–
robjohn♦Mar 29 '12 at 16:43

Okay thank you, One further question if you will: is $\theta(1,0,0) =0$ a fact?
–
EudenMar 29 '12 at 17:03

No, but since the plane of the motion is perpendicular to $(0,0,1)$ and the $0$-point for $\theta$ was not specified, I chose what seemed a reasonable direction for $\theta=0$ and made my assumption explicit.
–
robjohn♦Mar 29 '12 at 17:11

The angle from periapsis cannot be computed simply from the position and velocity of the orbiting body. The shape of the orbit depends on the strength of the gravitational field.
–
robjohn♦Mar 29 '12 at 20:32